Traveling wave solutions for the hyperbolic Cahn–Allen equation

Chaos, Solitons and Fractals 94 (2017) 75–79

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Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos

Traveling wave solutions for the hyperbolic Cahn–Allen equation I.G. Nizovtseva a,b,∗, P.K. Galenko a, D.V. Alexandrov b a b

Friedrich-Schiller-Universität Jena, Physikalisch-Astronomische Fakultät, Jena, D-07743, Germany Ural Federal University, Department of Mathematical Physics, Laboratory of Multi-Scale Mathematical Modeling, Ekaterinburg, 620000, Russian Federation

a r t i c l e

i n f o

Article history: Received 18 July 2016 Revised 31 October 2016 Accepted 22 November 2016

Keywords: Traveling wave Cahn–Allen equation First integral method Division theorem

a b s t r a c t Traveling wave solutions of the hyperbolic Cahn–Allen equation are obtained using the first integral method, which follows from well-known Hilbert–Nullstellensatz theorem. The obtained complete class of traveling waves consists of continual and singular solutions. Continual solutions are represented by tanh -profiles and singular solutions exhibit unbounded discontinuity at the origin of coordinate system. With the neglecting inertia of the dynamical system, the obtained traveling waves include the previous solutions for the parabolic Cahn–Allen equation.

1. Introduction The Cahn–Allen partial differential equation (CA-PDE) was suggested for the anti-phase boundary motion [1,2] and used in a wide spectrum of applications (see Ref. [3] and references therein). Being a useful tool within the phase-field models [4], the CA-PDE provides a framework for the mathematical description of freeboundary problems. One of important analytical solutions is related to traveling wave solutions (see Refs. [5,6] and references therein). Nowadays, one of convenient and complete ways in obtaining traveling waves lies in the use of the first integral method [7]. This method was introduced for a reliable treatment of the non-linear PDEs and over the last decades it has an intense period of its applicability [8,9]. This method can be considered as one of particular cases of the direct method [10] which generalizes the use of equivalent methods in finding exact solutions of PDE reduced to ODE [11]. Indeed, to date, several useful methods for obtaining solitons and traveling waves were developed. Among them, for instance, exist the tanh method [12], G /G–expansion method [13], and the other powerful method formulated as the rank analytical technique [14] applicable to a wide spectrum of nonlinear evolution PDEs [15]. To investigate some kinds of specific traveling waves, for example, solitary or periodic cusp waves, a phase–plane analysis seems to have a strong operability [16]. In the present work, we use the first integral method due to its evidence and simple applicability to parabolic and hyperbolic types of PDEs with the obtaining of complete set of traveling wave solutions [17,18].

© 2016 Elsevier Ltd. All rights reserved.

Special attention is given to the hyperbolic equations [19–23], especially, to CA-PDE with regard to its application in the field of fast phase transitions [23,24]. For the hyperbolic CA-PDE, a traveling wave solution in a particular form of tanh -function has been assumed [25,26]. So far, a general set of exactly obtained traveling waves for the hyperbolic CA-PDE is absent. Therefore, the main purpose of the present work is to find a general set of traveling waves in an exact analytical solution of the hyperbolic CA-PDE. The hyperbolic equation for the order parameter φ is given by [24]

τR

Corresponding author. E-mail address: [email protected] (I.G. Nizovtseva).

http://dx.doi.org/10.1016/j.chaos.2016.11.010 0960-0779/© 2016 Elsevier Ltd. All rights reserved.

(1)

where t is the time, D is the diffusion parameter for the order parameter φ , Mφ is the mobility of φ and τ R is the relaxation time for the gradient flow ∂ φ /∂ t. In the damped-wave Eq. (1) the inertial term τ R ∂ 2 φ /∂ t2 changes the type of the equation from usual dissipative parabolic type to the hyperbolic one with the drastic exchange of its analytical properties (as it has been shown for particular case of CA-PDE [27]). Physically reasonable applications of this equation are in processes with rapidly moving interfaces, in comparison with data of atomistic simulations on crystal growth kinetics, fast motion by mean curvature under large driving force of transformation between non-equilibrium states, inertial dynamics with dissipation, etc. (see the work [26] and references therein). With the free energy density f (φ ) = 14 (φ 2 − 1 )2 , Eq. (1) gives

τR ∗

∂ 2 φ ∂φ df (φ ) + = D∇ 2 φ − Mφ , ∂t dφ ∂t2

∂ 2 φ ∂φ + = D∇ 2 φ + Mφ (φ − φ 3 ), ∂t ∂t2

(2)

which is the hyperbolic CA-PDE. Introducing the dimesionless  relaxation time τ = τR Mφ , dimensionless coordinate x∗ = x Mφ /D,

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I.G. Nizovtseva et al. / Chaos, Solitons and Fractals 94 (2017) 75–79

dimensionless time t ∗ = Mφ t, and the new coordinate ξ = x∗ − ct ∗ which moves with the constant velocity, c = const, with the origin at φ = 1/2, Eq. (2) transforms to the one-dimensional ordinary differential equation (ODE)

d2 φ dφ (1 − τ c2 ) 2 + c + φ − φ 3 = 0. dξ dξ

of terms in determining the resulting number of unknown coefficients. Then, using the detailed form of Eq. (8) and having equating the coefficients at Y i (i = 0, 1, 2, 3 ), one gets

Y3 :

a˙ 2 (X ) = h(X )a2 (X ),

Y2 :

a˙ 1 (X ) = 2a2 (X )

Y1 :

a˙ 0 (X ) = 2a2 (X )

(3)

We shall use the first integral method [7] to check the existence of tanh -functions in the traveling waves of Eq. (3). 2. The first integral method and traveling-wave solutions Eq. (3) has the trivial constant solutions: +1, −1, and 0. For its non-constant solutions, Eq. (3) is considered as non-linear ODE     of the type of traveling wave solution Q φ , φ , φ , . . . = 0, where the prime denotes the derivative with respect to ξ . The solution of this ODE can be written in the form of φi (x∗ , t ∗ ) = F (ξ ), where i = 1, . . . , m. Now we introduce a new independent variable and its  derivative as X (ξ ) = F (ξ ) and Y (ξ ) = X (ξ ), respectively. According to the first integral method [7], X(ξ ) and Y(ξ ) are non-trivial solutions of Eq. (3):

dX (ξ ) = Y ( ξ ), dξ

(1 − τ c2 )

dY (ξ ) = X 3 (ξ ) − X (ξ ) − cY (ξ ), dξ

q [X ( ξ ) , Y ( ξ ) ] =

m 

ai (X )Y i = 0.

= [g(X ) + h(X )Y ]

m 

in the complex domain C(X, Y). From the latest expressions one obtains

i=0

2 

X3 − X = g(X )a0 (X ). 1 − τ c2

(16)

X4 X2 a0 ( X ) = − + 2 (1 − τ c2 ) 1 − τ c2 3 1 − τ c2 2 + A1 c + A1 2 2



1+





+X

c2 1 − τ c2

A0 c + A0 A1 1 − τ c2

 + d, (17)

2

ai (X )Y i .

(8)

The division theorem“ has been formulated in Ref. [7]. as a particular case of the Hilbert–Nullstellensatz theorem about characterization of maximal ideals in polynomial rings [28].

 3  A1 1 − τ c 2 = 0, 2

X4 :

2c +

X3 :

A0 1 − τ c 2 = 0,

X2 :

2c + 2A1 1 − τ c 2 + A1 c 2 +

i=0

A number of terms in Eqs. (7) and (8) are chosen as m = 2 to reach the required polynomial degree with the correct number 1

(15)

sult by (1 − τ c2 ) and having equating to zero the coefficients for different degrees of X, one gets

i=0

ai (X )Y i + h(X )Y

c X3 − X −2 + g(X )a1 (X ), 2 1 − τc 1 − τ c2

where d is the constant of integration. Substituting a0 (X) from Eq. (17) into Eq. (8), multiplying the re-

2 2  dai i+1  X 3 − X − cY Y + iaiY i−1 dX 1 − τ c2

= g( X )

(14)

Thus, we have found the expressions for the coefficients ai (X) from Eq. (6). In searching for solutions of Eq. (3), we assume the condition 1 − τ c2 > 0 which physically means that the interface velocity c cannot overcome and be larger than the maximum speed of disturbance propagation in the field of order parameter φ [24,25]. Such condition shrinks the set of all possible solutions (real and imaginary) to the class of real solutions. Solutions of Eq. (3) will be found for two possible cases: deg[g(X )] = 0 and deg[g(X )] = 1 which directly follow from Eqs. (8)–(12), see Appendix A.

(7)

where the coefficients ai for the solution (6) are

2 

c + g(X ), 1 − τ c2

Minding deg[g(X )] = 0 for Eqs. (13)–(16), one obtains g(X ) = A1 and a1 (X ) = 2cX/(1 − τ c2 ) + A1 X + A0 . With these expressions, the integral of Eq. (15) is

i=0

i=0

(13)

2.1. Case 0

ai (X )Y i ,

m m m ∂q  dai i  ∂Y  dai i = Y + iaiY i−1 = Y, ∂ X i=0 dX ∂ X dX i=0 i=0

a2 ( X ) = 1 ,

a1 ( X )

∂ q dX ∂ q dY ∂q ∂ q X 3 − X − cY + = Y+ ∂ X dξ ∂ Y dξ ∂X ∂ Y 1 − τ c2

(12)

where the point means the derivative d/dX. Since ai (X) are polynomials, from Eq. (9) it follows that a2 (X) ≡ const and h(X ) = 0. Then, accepting the value a2 (X ) = 1, Eqs. (9)–(12) are

a˙ 0 (X ) = a1 (X )

The polynomial (6) is known to be the first integral to Eqs. (4) and (5) due to the division theorem1 , if we suppose ai (X) to be polynomials of X and am (X) = 0. This first integral reduces Eq. (3) to a first order integrable ODE, which must have the exact analytical solutions. In Eq. (6), we consider X(ξ ) and Y(ξ ) as independent functions in the complex domain C(X, Y), therefore, dY/dX = 0. Following the division theorem [7,28], there exists the polynomial g(X ) + h(X )Y . Then, we shall write

(10)

(11)

X3 − X = g(X ) a0 (X ), 1 − τ c2

(5)

i=0

dq = dξ

a1 ( X )

Y0 :

a˙ 1 (X ) = 2

(6)

c + g(X )a2 (X ) + h(X )a1 (X ), 1 − τ c2

X − X3 c + a1 ( X ) 1 − τ c2 1 − τ c2 + g(X )a1 (X ) + h(X )a0 (X ),

(4)

expressed by X(ξ ) and Y(ξ ) using the following polynomial

(9)

+



(18)





(19)



2 1 1 − τ c2 A31 = 0, 2

 3 2  A1 c 1 − τ c 2 2 (20)

I.G. Nizovtseva et al. / Chaos, Solitons and Fractals 94 (2017) 75–79



Table 1 Sign counters Q, A and B for the set of solutions (25), (26), (40) and (41). The sign of the φ -profile velocity c by Eq. (23) shows concrete solution from the obtained set. The solution number, k = 1 . . . 8 Solution Solution Solution Solution Solution Solution Solution Solution

k=1 k=2 k=3 k=4 k=5 k=6 k=7 k=8







Q

A

B

c

+1 +1 +1 +1 −1 −1 −1 −1

+1 −1 +1 −1 +1 −1 +1 −1

+1 +1 −1 −1 +1 +1 −1 −1

> > > > < < < <





X1 :

A0. 1 − τ c 2 + A0 A1 c 1 − τ c 2 + 1 − τ c 2

X0 :

dA1 = 0.

2

+

+

0 0 0 0 0 0 0 0

X5 : X4 :

A A21 = 0, (21)

A1 = ± √ , 1 − τ c2

A0 = 0,

c=∓



2 1 + 9τ /2

,

d = 0.

2 + 9τ 4 2 + 9τ 2 X + X , 4 4



a1 (X ) = ∓ 2 + 9τ X.

For the coefficient a2 (X), we use again a2 (X ) = 1 due to h(X ) = 0. The obtained coefficients ai (X) together with Eq. (6) present finally the set of expressions for Yk by its quadratic equation which, using Eq. (4), gives

√

Yk (ξ ) =

 2 + 9τ  ±X (ξ ) ± X 2 (ξ ) , 2

k = 1 . . . 4.

(24)

Then, from Eq. (4), one can get the set of solutions for Xk (ξ ) as

Xk (ξ ) =

B + exp

√

A 2+9τ 2

ξ Q + c0

, k = 1 . . . 8,

(25)

where c0 is an arbitrary constant and the sign counters A, B are Q defined in Table 1. Taking the velocity c from Eq. (23) into account, we rewrite the solution for φ k (x, t) as



φk (x, t ) =

B + exp Q

A √

2+9τ 2

x∗ − 32 t ∗ + c0

, k = 1 . . . 8.

(26)

For Eqs. (13)–(16) one gets in the case deg[g(X )] = 1: g(X ) = A1 X + B0 and A1 = 0. Therefore, the coefficients ai (X) from the polynomial (6) should be defined by the different system of equations. Namely, Eq. (14) gives the expression a˙ 1 (X ) = 2 1−cτ c2 + A1 X + B0 , whose trivial integral is

c 1 X + A1 X 2 + B0 X + A0 . 2 1 − τ c2

(27)

Now, from Eq. (15), we obtain the expression a˙ 0 (X ) whose integral is

a0 ( X ) = −





1 1 + A21 X 4 + 8 2 (1 − τ c2 )



cA0 + A0 B0 X + d. 1 − τ c2

(28)

1 8

 = A31 ,

(29) 3B

5A2 c

5

0 2 1 2 + 2(1 − τ c2 ) = 8 A1 B0 + 6(1 − τ c2 ) , 2 1 − τc

(30)

A1 A0 5A1 B0 c A1 c 2 = −  + 2 2 2 1 − τc 6 (1 − τ c ) 2 1 − τ c2 1 − τ c2

X3 :

A1 + 32 A1 B0 c 1 + A1 B20 + A21 A0 , 2 1 − τ c2

+ X2 :

c2

2c

− +

2c

(31)

B c2

B

2B + 3B2 c

0 0 0 0 2 − 1 − τ c2 =  2 + 2(1 − τ c2 ) 2 2 1 − τc 1 − τc

1 3 1 cA0 A1 B + A1 A0 B0 + + A0 A1 B0 , 2 0 2 1 − τ c2

A0 cA0 B0 = + A0 B20 + A1 d, 1 − τ c2 1 − τ c2

X1 :

−

X0 :

dB0 = 0.

(32) (33) (34)

√  Eqs. (29) –(34) give d = 0 and A1 = ±2 2/ 1 − τ c2 . Multiply ing Eq. (33) by 1 − τ c2 , one gets A0 cB0 + B20 1 − τ c2 + 1 = 0,





that defines A0 = 0 with B20 1 − τ c2 + 4 > 0. Re-writing equations for the coefficients of the third and second powers of X and using B0 = 0, we obtain

√ √ 2 2 2 2 A0 = 0, A1 = ± √ , B0 = ± √ , 1 − τ c2 1 − τ c2 √ 3 2 c=±  , d = 0. 2 1 + 9τ /2

(35)

Now, using Eqs. (27), (28) and (35), the coefficients ai are

1 1 1  X 2, X4 − X3 +  2 (1 − τ c2 ) 1 − τ c2 2 1 − τ c2 √ √ 2 2 a1 ( X ) = ± √ X2 ∓ √ X, 1 − τ c2 1 − τ c2 a0 ( X ) =

√



2 + 3B0 c 1 1 + B20 + A1 A0 X 2 2 2 2 (1 − τ c2 )

since h(X ) = 0.

(36)

(37) (38)

Based on Eqs. (4) and (6), expressions (36)–(38) lead us to the quadratic equation with respect to Yk and to the final form of Xk function as

2.2. Case 1



1−τ

a2 ( X ) = 1,

Thus, we finalized the case deg[g(X )] = 0.

a1 ( X ) = 2

+

(1 − τ c2 )2

A1

(23)

In particular, the velocity c shows that the φ -profile can move in positive or negative direction with respect to the ξ -axis. Getting back to a0 (X), a1 (X) and a2 (X), from Eqs. (13)–(16), and taking into account c and A1 from Eq. (23), Eq. (17) gives

a0 ( X ) = −



0

As a result of the above treatments and using Eqs. (18)–(20), the whole set of parameters is given by

√ 3 2



c2

Substituting a1 (X) and a0 (X) from Eq. (27) and Eq. (28), respectively, into Eq. (17), we come to the following system for Xn

(22)

√ 2 2

77



5A1 c 1 + A1 B0 X 3 2 6 (1 − τ c2 )

Yk (ξ ) = Xk (ξ ) =

 2 + 9τ  ±X (ξ ) ± X 2 (ξ ) , 2

B + exp

√

A 2+9τ 2

ξ Q + c0

k = 1 . . . 4,

, k = 1 . . . 8,

(39) (40)

where c0 is an arbitrary constant and the sign counters A, B are Q defined in Table 1. Taking the velocity c from Eq. (23) into account, we rewrite the solution for φ k (x, t) as

φk (x, t ) =



B + exp Q

A √

2+9τ 2

x∗ − 32 t ∗ + c0

, k = 1 . . . 8.

Thus, the case deg[g(X )] = 1 has been considered.

(41)

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3. Discussion 3.1. Reduction to the solution of parabolic CA-equation With zero relaxation time, τ → 0, solutions (26) and (41), which has been obtained for the hyperbolic Eq. (2), transform to the traveling wave solutions



φk (x, t ) =

B + exp Q

A √

2 ∗ x 2

− 32 t ∗ + c0

,

k = 1 . . . 8,

(42)

for which sign counters A, B and Q are also defined in Table 1. Solutions (42) include itself solutions for the parabolic type of Allen– Cahn solution obtained earlier in Ref. [18,29].

set of four solutions, the order parameter tends to minus or plus infinity at the origin of the moving coordinate: φ k → ±∞ with k = 3, 4, 7, 8 at ξ = 0. These unbounded solutions can be eliminated from the complete set of possible solutions of the hyperbolic Cahn–Allen equation because, usually, analysis of analytical solutions is carried out for the set of bounded solutions. As one of examples, Bona and Schonbek [30] studied the existence and uniqueness of bounded traveling waves which tend to constant states far from the interface (which divides these two states). Additionally, from the physical viewpoint, the unbounded solutions have no meaning in tasks described by Cahn–Allen equation: the order parameter should have a smooth profile and be described by continuously differentiable function. Therefore, from the physical consideration, unbounded solutions should be omitted from the set of obtained solutions.

3.2. Specific solutions of the hyperbolic CA-equation 4. Conclusions 3.2.1. Bounded solutions In solutions (26) and (41) one can specially mark the class of traveling waves having the kink-profile. Namely, according to definition for the coefficients A, B and Q from Table 1, solutions 1, 2, 5 and 6 can be re-written with the zero value of the integration constant c0 = 0 in the following unified form:



φk (x, t ) =

i 1 − tanh 2



jξ



, k = 1, 2, 5, 6, with i = ±1,

δ

j = ±1,

(43)

for which the kink is moving in the direction or in the opposite direction of the coordinate ξ = x∗ − ct ∗ with the constant velocity c and with the characteristic width δ (having a meaning of correlation length of the phase field, see Refs. [25,26] and references therein):

c=∓



√ 3 2

2 1 + 9τ /2

,

δ=

√ 2 2

1 + 9τ /2

.

(44)

Expressions (43) and (44) represent tanh -profiles (described by the function of hyperbolic tangent) for the order parameter φ . Thus, we showed and confirmed that exact solutions for the hyperbolic Allen–Cahn Eq. (2) in a form of traveling wave can be described by the function of hyperbolic tangent that has been supposed in particular solutions of Refs. [25,26]. Whether the set of obtained tanh -functions (43) represents unique solutions of Eq. (3) – this task requires special consideration. Indeed, the general solution of Eq. (3) can be transformed to the standard Cauchy problem by means of the following substitution dφ /dξ = h(φ ):

( 1 − τ c 2 )h ( φ )

dh + ch(φ ) + φ − φ 3 = 0, h = h0 , dφ

φ = φ0 , (45)

where

ξ=

 φ dφ˜ , ξ = ξ0 , ˜) φ0 h ( φ

φ = φ0 ,

(46)

and subscript 0 designates the boundary values of the corresponding unknowns. As such, the uniqueness of the traveling waves described by Eqs. (43) and (44) might be verified by the analysis of the general solution (45) and (46), which can be done in a spirit of Bona and Schonbek [30]. With regard to the hyperbolic CA-PDE, a search for uniqueness solution might be provided in a class of real, continuous and differentiable functions (in R2 - or R3 -class). 3.2.2. Unbounded solutions Defining the coefficients A, B and Q from Table 1 for solutions 3, 4, 7 and 8, one can find √ that Eq. (26) and Eq. (41) have singularity at the point ξ = 2 + 9τ x∗ /2 − 3t ∗ /2 = 0. Indeed, for this

In the present work, Cahn–Allen equation of the hyperbolic type is considered. Using the first integral method [7], the soliton solutions as traveling waves have been analytically obtained for this equation. In their bounded form, these traveling waves are moving as tanh-profiles described by the function of hyperbolic tangent. This analytically confirms correctness of the particular solutions chosen in problems of rapid solidification [25,26]. The proven existence of traveling waves in a form of hyperbolic tangent function for the Cahn-Allen hyperbolic equation provides the ability to construct more complicated and much more rigorous analytical solutions of problems having an essential scientific merit and practical significance for the phase field crystal model [4,31]. In particular, using the amplitude wave representation [32], one may reduce the phase field crystal equation, which is sixth order in space, to the hyperbolic PDE, which has a form of advanced Cahn–Allen equation [33]. Application of the presently developed method to this advanced CA-PDE may give ability to find the complete set of solutions for amplitude equations with taking into account specific peculiarities of phase field crystals. Acknowledgments The authors acknowledge the support from the Russian Science Foundation (project no. 16-11-10095). I.N. specially acknowledges the support of Alexander von Humboldt Foundation (ID 1160779). P.G. specially acknowledges the support from German Research Foundation (DFG Project RE 1261/8-2). Appendix A. A choice of possible cases for the analysis For the presently analyzed Eq. (3) one can find only two cases. This follows from the definition of the g(X)-polynomial function, which should have the zero- or first-order by the X-argument. Indeed, in order to prove the sufficiency of the choice of the degrees for g(X) as 0 and 1, we have to balance the degrees of g(X), a0 (X), a1 (X). With this aim, let us denote deg(g(X )) = g0 , deg(g(a0 (X ))) = p0 , and deg(g(a1 (X ))) = p1 , where p0 , q1 , g0  Z. Then, from Eqs. (10)–(12), the following system is obtained

⎧ p1 − 1 = g0 , ⎪ ⎪ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎨3,

p0 − 1 = p1 , ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ p1 + p0 , ⎪ ⎩

p1 < 3 ∩ g0 + p1 ≤ 3 , p1 ≥ 3 ∩ 3 < g0 + p1 < p1 ,

(A.1)

p1 + p0 > 3 ∩ p1 + p0 > p1 ,

p1 + 3 = g0 + p0 .

Substituting the first expression of Eq. (A.1) into the third one, one gets p0 = 4. As a result, analyzing all three cases of the second equality of Eq. (A.1), we arrive at the following three cases.

I.G. Nizovtseva et al. / Chaos, Solitons and Fractals 94 (2017) 75–79

A) p0 = 4, 4 − 1 = 3, naturally. Because p0 , q1 , g0  Z and g0 = p1 − 1, one obtains p1 < 3 ∩ p1 − 1 + p1 ≤ 3, i.e., p1 < 3 ∩ p1 ≤ 2. Therefore, we have



p1 = 1, p1 = 2,

g0 = 0 , g0 = 1 .

(A.2)

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